Advertisements
Advertisements
Question
Evaluate the following limit:
`lim_(x -> 3)[sqrt(2x + 6)/x]`
Advertisements
Solution
`lim_(x -> 3)[sqrt(2x + 6)/x]`
= `(lim_(x -> 3) sqrt(2x + 6))/(lim_(x -> 3) x`
= `sqrt(2(3) + 6)/3`
= `sqrt(12)/3`
= `(2sqrt(3))/3`
APPEARS IN
RELATED QUESTIONS
Evaluate the following limit:
If `lim_(x -> 1)[(x^4 - 1)/(x - 1)]` = `lim_(x -> "a")[(x^3 - "a"^3)/(x - "a")]`, find all possible values of a
Evaluate the following limit :
If `lim_(x -> 5) [(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.
Evaluate the following limit :
`lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`
Evaluate the following limit :
`lim_(x -> 1) [(x + x^3 + x^5 + ... + x^(2"n" - 1) - "n")/(x - 1)]`
Evaluate the following :
`lim_(x -> 0)[x/(|x| + x^2)]`
Evaluate the following :
`lim_(x -> 0) [(sqrt(1 - cosx))/x]`
Evaluate the following :
`lim_(x -> 1) [(x + 3x^2 + 5x^3 + ... + (2"n" - 1)x^"n" - "n"^2)/(x - 1)]`
Evaluate the following :
`lim_(x -> 0) {1/x^12 [1 - cos(x^2/2) - cos(x^4/4) + cos(x^2/2) cos(x^4/4)]}`
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 2) (x - 2)/(x^2 - 4)`
| x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
| f(x) | 0.25641 | 0.25062 | 0.250062 | 0.24993 | 0.24937 | 0.24390 |
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) (sqrt(x + 3) - sqrt(3))/x`
| x | – 0.1 | – 0.01 | – 0.001 | 0.001 | 0.01 | 0.1 |
| f(x) | 0.2911 | 0.2891 | 0.2886 | 0.2886 | 0.2885 | 0.28631 |
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 1) (x^2 + 2)`
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 1) f(x)` where `f(x) = {{:(x^2 + 2",", x ≠ 1),(1",", x = 1):}`
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 3) 1/(x - 3)`
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 5) |x - 5|/(x - 5)`
If f(2) = 4, can you conclude anything about the limit of f(x) as x approaches 2?
Evaluate the following limits:
`lim_(sqrt(x) -> 3) (x^2 - 81)/(sqrt(x) - 3)`
Evaluate the following limits:
`lim_(x -> 1) (sqrt(x) - x^2)/(1 - sqrt(x))`
Evaluate the following limits:
`lim_(x -> 1) (root(3)(7 + x^3) - sqrt(3 + x^2))/(x - 1)`
Show that `lim_("n" -> oo) (1^2 + 2^2 + ... + (3"n")^2)/((1 + 2 + ... + 5"n")(2"n" + 3)) = 9/25`
Show that `lim_("n" -> oo) 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/("n"("n" + 1))` = 1
Evaluate the following limits:
`lim_(x -> oo)(1 + 1/x)^(7x)`
Evaluate the following limits:
`lim_(x -> oo)(1 + "k"/x)^("m"/x)`
Evaluate the following limits:
`lim_(x -> 0) (sin^3(x/2))/x^2`
Evaluate the following limits:
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m"`
Evaluate the following limits:
`lim_(x -> 0) (1 - cos^2x)/(x sin2x)`
Evaluate the following limits:
`lim_(x -> pi) (sin3x)/(sin2x)`
Evaluate the following limits:
`lim_(x -> 0) ("e"^("a"x) - "e"^("b"x))/x`
Evaluate the following limits:
`lim_(x -> 0) (tan x - sin x)/x^3`
Choose the correct alternative:
`lim_(theta -> 0) (sinsqrt(theta))/(sqrt(sin theta)`
Choose the correct alternative:
`lim_(x -> 0) ("a"^x - "b"^x)/x` =
Choose the correct alternative:
`lim_(x -> 3) [x]` =
Choose the correct alternative:
`lim_(alpha - pi/4) (sin alpha - cos alpha)/(alpha - pi/4)` is
Choose the correct alternative:
`lim_(x -> 0) ("e"^tanx - "e"^x)/(tan x - x)` =
`lim_(x -> 0) (sin 4x + sin 2x)/(sin5x - sin3x)` = ______.
`lim_(x→-1) (x^3 - 2x - 1)/(x^5 - 2x - 1)` = ______.
